When are the units of R[x] exactly the units of R?

I (Anton) have edited this question to be the question Pete and Zeb discuss in the first few comments.

What conditions on a ring $R$ imply that the units of $R[x]$ are exactly the units of $R$?

• As with your last question, this is an undergraduate level homework-type question, hence not what Math Overflow is designed for. (Also, please don't use the "math-education" tag as a synonym for "Help me with my math questions".) But here's a hint: for any commutative ring R, show that the units of R[x] are precisely the units of R. – Pete L. Clark Mar 2 '10 at 3:05
• Your hint is kind of false. Let R = k[e]/e^2, then in R[X] we have (1+eX)(1-eX) = 1-e^2X^2 = 1, and 1+eX, 1-eX are not in R. The hint becomes true if the ring R is an integral domain. (Does anyone know if there is a weaker condition we can impose on R making the hint true?) – zeb Mar 2 '10 at 3:09
• @ZC: Thanks! I should have said integral domain (e.g. Z!). Also, it is very tempting for me to edit the question to put your last question into it: it would make for a much better MO question. – Pete L. Clark Mar 2 '10 at 3:13
• P.S.: "Kind of false"?!? What, is that supposed to make me feel better? :) – Pete L. Clark Mar 2 '10 at 3:15
• If you're using commutative rings with unity, then you can just require the ring is reduced; see exercise I.2 in Atiyah & Macdonald. – Steve D Mar 2 '10 at 3:34

If $R$ is a commutative ring, then by the following result, the answer is "if and only if $R$ is reduced."

If $R$ is a commutative ring, then $a_0+a_1x+\cdots + a_nx^n\in R[x]$ is a unit if and only if $a_0$ a unit in $R$ and $a_i$ is nilpotent for $i>0$.

Proof. One direction is easy. Any polynomial of the given form is a unit because the sum of a unit and a nilpotent element is always a unit.

The other direction isn't too hard if $R$ is a domain (the product of non-zero elements is always non-zero). If $g=b_0+\cdots b_mx^m$ (with $b_m\neq 0$) is the inverse of $f=a_0+\cdots+a_nx^n$ (with $a_n\neq 0$), then the highest order term of $1=f\cdot g$ is $a_nb_mx^{n+m}$, so we must have $n=m=0$ and $a_0$ invertible (with inverse $b_0$)

For the general case, suppose $a_0+\cdots +a_nx^n$ is a unit. Reducing modulo $x$, we must get a unit in $R[x]/(x)\cong R$, so $a_0$ must be a unit. Reducing modulo any prime $\mathfrak p\subseteq R$, we get a unit in $(R/\mathfrak p)[x]$. Since $R/\mathfrak p$ is a domain, the previous paragraph shows that $a_i\in \mathfrak p$ for all $i>0$ and all primes $\mathfrak p$. Since the intersection of all primes is the nilradical, each $a_i$ must be nilpotent.

A more "bare hands" elementary proof is given in Ex. 1.32 of Lam's Exercises in Classical Ring Theory. He also gives counterexamples to both implications if $R$ is not assumed commutative and mentions a really interesting related question. If $I\subseteq R$ is an ideal all of whose elements are nilpotent and $a_i\in I$, then does it follow that $1+a_1x+\cdots +a_nx^n$ is a unit in $R[x]$? If you can prove that it does, it would imply the Köthe conjecture, a famous problem in ring theory.

• +1: way to make lemonade. – Pete L. Clark Mar 2 '10 at 6:00